RI DHS HCI TMJC 9649 2023 Prelim P1 Solutions

FM Prelim Paper 1 (100 marks) 1(a) [2] Let 32f ( ) e 2 xxx    3 2(1) 3 2(2) f (1) (1) e 2 1.1353 (5 s.f.) < 0 f (2) (2) e 2 5.9817 (5 s.f.) > 0           [Alternatively, since 0 < 2e < 1 and 0 < 4e <1, 3 2(1) 2 3 2(2) 4 f (1) (1) e 2 1 e < 0 f (2) (2) e 2 6 e > 0]             Since 22f '( ) 3 +2e 0 xx x x     , f is strictly increasing and continuous. Hence there is exactly one root,  , that lies in [1,2]. (b) [2] Let 32f ( ) e 2 xxx    Using linear interpolation on [1,2] , we have 1 (1)f (2) (2)f (1) 1.1595 1.2 (1 d.p.)f (2) f (1)     (c) [2] 22f '( ) 3 +2e xxx  By Newton-Raphson method, we have 23 1 22 f( ) e 2 f ( ) 3 +2e n n nn n n n n n               Take the initial approximation to be 1 1.2  , 2 1.28058  , 3 1.27609  , 4 1.27607  Checking f (1.2755) 0.00289 < 0 f (1.2765) 0.00215 > 0   That is, 1.276 (3 d.p.)  . 2(a) [2] Sequence converges  as n , nul  and 1nul  8 21 2 ll l   l satisfies the equation 8 21 2 x xx   ,

2 8 21 2 ( 2) 8 21 10 21 0 3 or 7 x xx x x x xx xx          Hence the sequence converges to either 3 or 7. (proven) (b) [4]   1 8 2177 2 8 21 7 2 2 7 0 ( 7 0, 2 0)2 n n n nn n n nn n uu u uu u u uuu                1 7nu  Next, consider 1nnuu  ,   1 8 21 2 8 21 2 2 n n n n n n n n n uu u u u u u u u          2 10 21 2 nn n uu u     37 2 nn n uu u   Since 2nu  , 20nu  . Since 37 nu ,   3 7 0nnuu   . Hence    1 1 37 0 2 0 for 3 7 nn nn n n n n uuuu u u u u            Thus 1nnuu  if 37 nu . Thus, 1 7nnuu  if 37 nu .

(c) [2] From the graph, the sequence strictly increases and converges to 7. Alternatively, Apply (b) repeatedly. 1 1 2 2 23 3 34 3 7 7 37 7 37 7 ... u u u u uu u uu                   Hence we have 1 2 3 4 5 13 ... ... 7 nnu u u u u u u           , and so the sequence strictly increases and converges to 7. 3(a) [3] The characteristic equation for 212 2 0n n nax a x x   is 22 2 1 0aa   , and so for 0a , 2 4 4(2 )(1) 2 ( 2 distinct roots )2(2 ) 2 a a a a a a aa     412 = 1 e22 22 ia a a i a iaa aa       Hence the general solution is 11 22 nn n iix A B aa            .

or 2 ( )cos ( ) sin 442 n n nnx A B A B i a                  (note that A and B are complex numbers) or 2 cos sin , ,442 n n nnx C D C D a                 (b) [3] When 3a , the general solution is 2 cos sin , ,4423 n n nnx C D C D                 0 33xC   1 1 2 3 1 1 3 2 1 3 2 3 2 2 3 x D D D          Hence 1 3cos sin446 n n nnx               , and so 3, 1,CD  1f ( ) , g( ) 46 n nnn  . Alternatively When 3a , the general solution is 11 2 3 2 3 nn n iix A B            0 33x A B            1 11 1 ( ) 2 3 2 3 2 3 1 1 13 ( ) 3 ( ) 2 2 3 3 ABx i i A B A B i A B i A B i A B i i                        113 , 322A i B i     Hence

44 3 1 3 1 22 2 3 2 3 3 2 3 2 22 2 3 2 3 3 cos sin442( 6) 3 cos sin442( 6) nn n nn ii n n i i i ix i e i e i n n i i n n i                                                                  1 6cos 2sin442( 6) 1 3cos sin44( 6) n n nn nn                             (c) [2] For 0a , The general solution is 2 cos sin , ,442 n n nnx C D C D a                 Note that the trigonometric terms are periodic and bounded. Hence for 0nx  , we only require 2 1 1 10 1 , 2222 aa a          is the required range. 4(a) [3] Since 1 2 3 4 0       , the eigenvalues are distinct and so { 1 2 3 4, , ,x x x x } is linearly independent. { 1 2 3 4, , ,x x x x } linearly independent  { 1 2 3 4, , ,x x x x } is a basis of 4 . Hence for any 4x , it can be expressed as a linear combination of 1 2 3 4, , ,x x x x 1 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 1 1 2 2 3 3 4 4 1 1 1 2 2 2 3 3 3 4 4 4 c c c c c c c c c c c c c c c c                     x x x x x Ax A x A x A x A x Ax Ax Ax Ax x x x x By pre-multiplying by A repeatedly, we have

2 1 1 1 2 2 2 3 3 3 4 4 4 1 1 1 2 2 2 3 3 3 4 4 4 2 2 2 2 1 1 1 2 2 2 3 3 3 4 4 4 3 3 3 3 3 1 1 1 2 2 2 3 3 3 4 4 4 1 1 1 2 2 2 3 3 3 4 4 4 A x A x A x A x A x Ax Ax Ax Ax x x x x A x x x x x A x x x x xk k k k k c c c c c c c c c c c c c c c c c c c c                                         By factorizing out 1 k , we have 324 1 1 1 2 2 3 3 4 4 111 A x x x x x kkk kk c c c c                         (b) [1] From (a), since 1 2 3 4 0,       we observe that for large k , 1 0, 2,3, 4 k i i     , since 1 1i   , and so 324 1 1 1 2 2 3 3 4 4 1 1 1 111 A x x x x x x kkk k k k c c c c c                          . (c) [1] Using the GC to solve for Mx = –x as a SLE, we get the eigenvector associated with eigenvalue 1 as 1 1 0 1       . (d) [3] 1 2 3 41.80194, 1.24698, 1, 0.44504,       Note that the eigenvalues of M are all distinct, so results in (a) and (b) holds. Note that it suffices to find the least integer value of k by considering 2 1 0.0001 k      as 324 111 k k k                    .From GC, k 2 1 0.0001 k      25 0.0001 26 0.00007 < 0.0001 Thus least k = 26

Since 26 26 11 1557649 1942071 1942071 864201 c       x M x and 26 1c is a constant, x1 can be scaled accordingly, and so an approximate eigenvector x1 is 0.80 1 1 0.44       . 5(a) [3] det () AI 2 2 2 2 2 2( )( ) ( )( ) ( ) ( )a a b a b a a b a b               . Hence det () AI 0 22 0bb     Eigenvalues of A satisfy the characteristic equation 22 0 b which has roots b . Hence b is an eigenvalue of A. [Alternatively, Since the column sums of A are both b, AT 1 1   = b 1 1   , and hence b is an eigenvalue of AT with eigenvector 1 1   . But eigenvalue of A is same as eigenvector of AT . Hence b is an eigenvalue of A.] To find corresponding eigenvector, solve: 0 0 a b b a x b a a b y                 Since 21 00 RRa b b a a b b a b a a b                  , bayx ba   , and so an eigenvector corresponding to b is 1 ba ba     or    ba ba . [Alternatively, ( ) 0, xa b b a y     and so an eigenvector corresponding to b is perpendicular to    ab ba which is    ba ba .]

(b) [3] 2 2 2 2 2 2 2 2 2 222 2 00 00 10 01 A IA a b a a b a b a a b a a a b a b b a a b bb                          22 2 22 1 1 2 1 1 IA IAA A b b a b a a b a b a a b a a b                   OR directly             1 2 2 2 2 1 det 1 ( )( ) 1 () a b a b a a a b a b a aa b a b a a b a b a aa b a                     A A = 2 1 a b a b a ab    (c) [5] 2 222 22 2 3 13 2 2 44 2 4 5 15 100 010 a b a b bbb a a b a b abb b a a bb bb                           A I A AA AIA AAA Hence 1 2 if is odd if is even n n n bn b